1. Nov 3, 2003

### Juntao

Consider the function z=f(x,y). If you start at the point (4,5) and move toward the point (5,6), the direction derivative is sqrt(2). Starting at (4,5) and moving toward (6,6), the directional derivative is sqrt(5). Find gradient f at (4,5).

Okay, this is probably a simple problem, but I don't know how to start it. Help appreciated.

2. Nov 3, 2003

### Hurkyl

Staff Emeritus
Do you know a formula that relates gradients and directional derivatives?

What do you get when you plug what you know into that formula? (represent unknowns with variables)

3. Nov 3, 2003

### Juntao

Um, I think this is the one, hopefully:
The derivative of f at Po in the direction of u= gradient f dot with u = magnitude grad f * magnitude u * cos theta = magnitude grad f * cos theta.

Or like D_u f= grad f (dot) u = |grad f| |u| cos theta = |grad f| cos theta

Maybe that'll help me out?

4. Nov 4, 2003

### HallsofIvy

Staff Emeritus
Yes, that's true as long as u is a unit vector (has length 1).

Another way of saying this is that the directional derivative is
f_x u_x+ f_y u_y where f_x is the partial derivative and u_x is the x component of unit vector u.

Find unit vectors in both the given directions and write out
f_x u_x+ f_y u_y= &radic;(2) and

That gives you two equations for f_x and f_y.

5. Nov 4, 2003

### Hurkyl

Staff Emeritus
That's the one of which I was thinking.

You know two instances of D_u f and u. (don't forget that u is a unit vector!), so grad f is the unknown. You have two equations, so you should be able to solve for the two components of grad f.